Simplified form for a Newton's sum? $a_1^k + a_2 ^k + \ldots + a_n^k = k $ for $k=1,2,\ldots,n$. Let $ a_1, a_2, a_3, \ldots , a_n $ be complex number satisfying $ \displaystyle \sum_{j=1}^n a_j ^k= k $ where $ k =1,2,\ldots, n $. 
Prove (or disprove) that $\displaystyle \sum_{j=1}^n a_j ^{n+1} $ can be expressed $ \dfrac{K}{n!} $, where $K$ is an integer satisfying $ \gcd(K, n!) = 1 $.

I've made up this question. I have shown that it is true for $n=1,2,3,4$. I'm not sure whether it holds for all positive integer $n$.
 A: I will use the notation and the concepts outlined in this question of mine.
To recover the symmetric elementary polynomials of $a_1,a_2,\ldots,a_n$ it is enough to apply $\text{EXP}$:
$$ \sum_{r\geq 0}(-1)^r e_r x^r = \exp\left(-\sum_{m\geq 1}\frac{p_m}{m} x^m\right) $$
hence the elementary symmetric polynomials are given by the coefficients of the Taylor series of $\exp\left(\frac{x}{x-1}\right)$. $p_{n+1}$ can be recovered from $p_1,p_2,\ldots,p_n$ by $CH$ - we have
$$ x^n = e_1 x^{n-1} - e_2 x^{n-2} +\ldots+ (-1)^{n+1} e_n $$
for any $x=a_i$, hence by multiplying both sides of the previous expression by $x$ and summing over $x=a_1,a_2,\ldots,a_n$ we get:
$$\begin{eqnarray*} p_{n+1} &=& e_1 p_n - e_2 p_{n-1} + \ldots + (-1)^{n+1} p_1\\&=&n e_1-(n-1)e_2+\ldots+(-1)^{n+1} \end{eqnarray*}$$
and your conjecture holds, since such expression can be recovered from the derivative of a partial sum of the Taylor series of $\exp\left(\frac{x}{x-1}\right)$, which has the following property:
$$ \exp\left(\frac{x}{x-1}\right) = \sum_{n\geq 0}\frac{K_n}{n!} x^n,\qquad K_n\in\mathbb{Z}, \,(K_n,n!)=1.$$
